我试图建立一个从在线MySQL数据库使用数据的iPhone应用程序。 (请注意,编程是新的我)iPhone应用程序 - Mysql数据库连接不工作
我有这样一段代码在我的应用程序,这是应该连接到我的PHP文件:
//loginAction
- (void) loginAction
{
if ([email.text isEqualToString:@""] || [password.text isEqualToString:@""]) {
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error!" message:@"please fill in everything":self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
[alert show];
return;
}
//database connection
NSString *strURL = [NSString stringWithFormat:@"http://db.imagine-app.nl/login.php?email=%@&password=%@", email.text, password.text];
//execute php
NSData *dataURL = [NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];
//receive data
NSString *strResult = [[NSString alloc] initWithData:dataURL encoding:NSUTF8StringEncoding];
if ([strResult isEqualToString:@"1"])
{
action:@selector(sendLogin);
}else
{
//invalid information
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error" message:@"E-mail or password wrong" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
[alert show];
return;
}
}
在我的网站(由Web托管公司托管)我有一个名为“login.php中”一个php文件,其中包含下面的代码:
<?php
if (isset($_GET["email"]) && isset($_GET["password"])){
$email = $_GET["email"];
$password = $_GET["password"];
$result = login($email, $password);
echo $result;
}
function makeSqlConnection()
{
$DB_HostName = "db.imagine-app.nl";
$DB_Name = "*************";
$DB_User = "*************";
$DB_Pass = "*************";
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die(mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
return $con;
}
function disconnectSqlConnection($con)
{
mysql_close($con);
}
function login($email, $password)
{
$con = makeSqlConnection();
$sql = "SELECT * from user WHERE email = '$email' AND password = '$password';";
$res = mysql_query($sql,$con) or die(mysql_error());
$res1 = mysql_num_rows($res);
disconnectSqlConnection($con);
if ($res1 != 0) {
return 1;
}else{
return 0;
}// end else
}
?>
代码这两件应让我来访问我的数据库,但是当我测试它(把电子邮件和密码在表,并尝试从我的应用程序登录),iPhone模拟器stuses一段时间,之后,我告诉米Ë我输入的值是错误的(这意味着该查询返回值。) 以何种方式做我必须编辑这些两段代码,使其工作? 预先感谢您。
1st close the'''here:'NSString * strURL = [NSString stringWithFormat:@“http://db.imagine-app.nl/login.php?email=%@&password=%@”,email。文本,password.text];' – 2013-03-18 21:01:19